Question 1.
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
i. 135 and 225
As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,
225 = 135 × 1 + 90
Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,
135 = 90 × 1 + 45
Again, 45 ≠ 0, repeating the above step for 45, we get,
90 = 45 × 2 + 0
The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.
Hence, the HCF of 225 and 135 is 45.
ii. 196 and 38220
In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,
38220 = 196 × 195 + 0
We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.
Hence, the HCF of 196 and 38220 is 196.
iii. 867 and 255
As we know, 867 is greater than 255. Let us apply now Euclid’s division algorithm on 867, to get,
867 = 255 × 3 + 102
Remainder 102 ≠ 0, therefore taking 255 as divisor and applying the division lemma method, we get,
255 = 102 × 2 + 51
Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,
102 = 51 × 2 + 0
The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.
Hence, the HCF of 867 and 255 is 51.
Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some
integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0 ≤ r < 6.
Now substituting the value of r, we get,
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5, respectively.
If a = 6q, 6q + 2, 6q + 4, then a is an even number and divisible by 2. A positive integer can be either even or odd,
Therefore, any positive odd integer is of the form of 6q + 1, 6q + 3, and 6q + 5, where q is some integer.
Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
Given,
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, 32), gives the maximum number of columns in which they can march.
By Using Euclid’s algorithm to find their HCF, we get,
Since, 616 > 32, therefore,
616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as new divisor, we have,
32 = 8 × 4 + 0
Now we have got remainder as 0, therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.
Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Solution:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, x = 3q, 3q + 1 and 3q + 2
Now as per the question given, by squaring both the sides, we get,
x2 = (3q)2 = 9q2 = 3 × 3q2
Let 3q2 = m
Therefore, x2 = 3m ……………………..(1)
x2 = (3q + 1)2 = (3q)2 +12 +2 × 3q × 1 = 9q2 + 1 +6q = 3 (3q2 +2q) +1
Substitute, 3q2 +2q = m, to get,
x2 = 3m + 1 ……………………………. (2)
x2 = (3q + 2)2 = (3q)2 + 22 + 2 × 3q × 2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1
Again, substitute, 3q2 + 4q + 1 = m, to get,
x2 = 3m + 1…………………………… (3)
Hence, from equations 1, 2, and 3, we can say that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Question 5.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q + r, where q ≥ 0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, putting the value of r, we get,
x = 3q
or
x = 3q + 1
or
x = 3q + 2
Now, by taking the cube of all the three above expressions, we get,
Case (i): When r = 0, then,
x2 = (3q)3 = 27q3 = 9(3q3)= 9m; where m = 3q3
Case (ii): When r = 1, then,
x3 = (3q+1)3 = (3q)3 +13 +3 × 3q × 1 (3q + 1) = 27q3 + 1 + 27q2 + 9q
Taking 9 as common factor, we get,
x3 = 9(3q3 +3q2 +q)+1
Putting = m, we get,
Putting (3q3 +3q2 + q) = m, we get ,
x3 = 9m+1
Case (iii): When r = 2, then,
x3 = (3q+2)3 = (3q)3 + 23 +3 × 3q × 2 (3q + 2) = 27q3 + 54q2 + 36q + 8
Taking 9 as common factor, we get,
x3 =9(3q3 +6q2 +4q)+8
Putting (3q3 +6q2 +4q) = m, we get,
x3 = 9m+8
Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Class 10 Maths Real Numbers
Rational numbers and irrational numbers are taken together form the set of real numbers. The set of real numbers is denoted by R. Thus every real number is either a rational number or an irrational number. In either case, it has a non–terminating decimal representation. In the case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is non–repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line ‘l’ or we may say that every point on the line ‘l’ corresponds to a real number (rational or irrational).
From the above discussion we may conclude that:
To every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number. Thus we see that there is one–to–one correspondence between the real numbers and points on the number line ‘l’, that is why the number line is called the ‘real number line’.
Chapter 1 Class 10 Maths Real Numbers Summary
We have studied the following points:
1. Euclid’s Division Lemma: Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r where 0 = r = b.
2. Euclid’s Division Algorithm: According to this, which is based on Euclid’s division lemma, the HCF of any two positive integers a and b with a > b is obtained as follows:
Step 1 Apply the division lemma to find q and r where a = bq + r, O = r < b.
Step 2 If r = 0, the HCF is b . If r? 0 apply Euclid Lemma to b and r
Step 3 Continue the process until the remainder is zero. The divisor at this stage will be HCF (a, b). Also HCF (a, b) = HCF (b, r)
3. The Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorized) as a product of primes and this factorization is unique, apart from the order in which the prime factors occur.